I am going to ask a simple question, for sure.
The pressure with respect to the altitude is given by this formula
Where
- sea level standard atmospheric pressure p0 = 101325 Pa
- sea level standard temperature T0 = 288.15 K
- Earth-surface gravitational acceleration g = 9.80665 m/s2.
- temperature lapse rate L = 0.0065 K/m
- universal gas constant R = 8.31447 J/(mol·K)
- molar mass of dry air M = 0.0289644 kg/mol
( from Wikipedia )
In addition to this, we have $L = \frac{g}{C_p}$ where $g = 9.80665\ m/s^2$ and $C_p$ is the constant pressure specific heat $ = 1007\ J/(Kg\ K)$
Understood and thence the above formula can be written in this simple way:
$$\boxed{P = P_0\left(1 – \alpha h\right)^{\beta}}$$
where
$\alpha = \frac{g}{C_p\ T_0} \approx 3.3796\cdot 10^{-5}\ m^{-1}$
$\beta = \frac{C_p\ M}{R} \approx 3.5081971$
On the other side we learnt from the elementary physics that the formula for the pressure is also given by
$$\boxed{P = P_0 + \rho g h}$$
Where $\rho$ is the air density $(1.23\ kg/m^3)$.
The question
First thing first: I assumed that
$$h = h_1 – h_0$$
is that correct? I mean it's the difference between two heights (maybe from a table and the ground, just to say).
Since the two formulae seems quite different, I tried with a numerical example in calculating the pressure in two points, with a difference of height about $0.18$ m and I got a really similar result.
Since the first formula is more technique, I think it's the correct formula but I would like to understand if one could pass from the first to the second or vice versa somehow.
Also I would like to know if there are cases in which I can use only the second formula or only the first formula!
Best Answer
For a compressible gas, your elementary physics formula changes to: $$\frac{dp}{dz}=-\rho g$$whre z is the elevation (above ground level) and $\rho$ is the density of the gas. From the ideal gas law, $$\rho=\frac{pM}{RT}$$ where M is the molecular weight. If we combine these two equations, we get the "barotropic" equation:$$\frac{1}{p}\frac{dp}{dz}=-\frac{Mg}{RT}$$If we integrate the barotropic equation from 0 to z, we get:$$p=p_0\exp\left(-\frac{Mg}{R}\int_0^zT(z')dz'\right)$$This is leading to your first equation.